Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Shear force & bending moment

Shear and moment: V = ΣF on one side, M = ΣF·d on one side
Bending stress: σ = M·y/I
Rectangle: I = bh³/12
Section modulus: W = I/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid: ȳ = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Beam deflection

Point load at midspan: δ = PL³/(48EI)
Uniform load, simply supported: δ = 5wL⁴/(384EI)
Cantilever, end load: δ = PL³/(3EI)
Cantilever, uniform load: δ = wL⁴/(8EI)
Superposition: δ = δ₁ + δ₂

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 300 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: copper, Ø60 mm, 300 mm, E = 120 GPaδ = 0.5 mm

P = kN

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 115 kNsegment 1: Ø30 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 1.5 m, brass (E = 100 GPa)

δ = mm

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 1500 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

T = 1500 N·md_i/d_o = 0.7τ_allow = 50 MPa

d_o = mm

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 300 rpmd_o = 80 mmd_i = 64 mm

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 20 kN·m at x = 5 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 10 mM₀ = 20 kN·m clockwise, at 5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 40 kN at 2 mw = 6 kN/m from 0 to 8 mx = 7 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 120 mm × 150 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 60 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 150 mmd = 50 mmhole centre: y = 60 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 300 mm plate has a rectangular cut-out 80 mm wide and 20 mm high, centred on the vertical centre line with its centre 120 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 300 mmcut-out: 80 × 20 mmhole centre: y = 120 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 0 MPa, σᵧ = 100 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 0 MPaσ_y = 100 MPaτ_xy = 50 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 400 mm and wall thickness 6 mm holds gas at a gauge pressure of 1 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 400 mmt = 6 mmp = 1 MPaα = 55°

σ_n = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 20 kN/mL = 2.5 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

θ_A = rad

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 20 kN/mL = 2 mE = 11 GPa (timber)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 2 mE = 70 GPa (aluminium)b × h = 25 × 50 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 5 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 47.92 kN

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A₁ = 314.2 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·314.2) = 9.549 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(120000·2827) = 8.842 × 10⁻⁷ mm/N

Load

P = 0.5/1.043 × 10⁻⁵ = 47923 N = 47.92 kN

Answer

P = 47.92 kN

2.Axial stress & strain

Answer: δ = 12.07 mm

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 162.7 MPa

σ₂ = 572 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 162.7·1500/70000 = 3.486 mm

δ₂ = σ₂L₂/E₂ = 572·1500/100000 = 8.579 mm

δ = δ₁ + δ₂ = 12.07 mm

Answer

δ = 12.07 mm

3.Torsion of shafts

Answer: d_o = 58.58 mm

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Outer diameter from the allowable stress

T = 1500 N·m = 1.5 × 10⁶ N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·1.5 × 10⁶/(π·50·0.7599)) = 58.58 mm

d_i = 0.7·58.58 = 41.01 mm

Answer

d_o = 58.58 mm

A solid shaft would need d = 53.46 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: τ_max = 10.73 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 20000/31.42 = 636.6 N·m

Polar second moment of area

J = π(80⁴ − 64⁴)/32 = 2.374 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 636620·40/2.374 × 10⁶ = 10.73 MPa

Answer

τ_max = 10.73 MPa

5.Shear force & bending moment

Answer: M_x = −10 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 20/10 = 2 kN (upward)

R_A = −R_B = −2 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −2·5 = −10 kN·m

Answer

M_x = −10 kN·m

6.Shear force & bending moment

Answer: M_x = 31 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·8 = 48 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 40·2 + 6·8·4

R_B = 272/8 = 34 kN

ΣF_y = 0: R_A = 88 − 34 = 54 kN

Cut at x = 7 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 54·7 − 40·(7 − 2) − 6·7·3.5 = 31 kN·m

Answer

M_x = 31 kN·m

7.Centroid & second moment of area

Answer: I_x = 32.95 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (18000·75 − 1963·60)/(18000 − 1963) = 76.84 mm

Full plate about ȳ

120·150³/12 + 18000·−1.837² = 33.81 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·−16.84² = 0.8634 × 10⁶ mm⁴

Subtract

I_x = 32.95 × 10⁶ mm⁴

Answer

I_x = 32.95 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 448.5 × 10⁶ mm⁴

Centroid

ȳ = (60000·150 − 1600·120)/(60000 − 1600) = 150.8 mm

Full plate about ȳ

200·300³/12 + 60000·−0.8219² = 450 × 10⁶ mm⁴

Hole about ȳ

80·20³/12 + 1600·−30.82² = 1.573 × 10⁶ mm⁴

Subtract

I_x = 448.5 × 10⁶ mm⁴

Answer

I_x = 448.5 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = 22.5 °

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = 50/70.71 = 0.7071, sin 2θ_s = −(−50)/70.71 = 0.7071

2θ_s = atan2(50, 50) = 45°

Angle on the element

θ_s = 22.5°

check: θ_s − θ_p = 22.5 − 67.5 = −45°

Answer

θ_s = 22.5°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_n = 22.15 MPa

Wall stresses

r = 400/2 = 200 mm

σ_h = pr/t = 1·200/6 = 33.33 MPa

σ_a = pr/(2t) = 16.67 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 16.67·sin²55° + 33.33·cos²55° = 22.15 MPa

Answer

σ_n = 22.15 MPa

11.Beam deflection

Answer: θ_A = 0.007659 rad

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

w = 20 N/mm, L = 2500 mm

EI = 1.7 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 20·2500³/(24·1.7 × 10¹²) = 0.007659 rad

Answer

θ_A = 0.007659 rad

That is 0.4388°.

12.Beam deflection

Answer: I_req = 68.18 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 2000/360 = 5.556 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·20·2000⁴/(384·11000·5.556) = 6.818 × 10⁷ mm⁴

I_req = 68.18 × 10⁶ mm⁴

Answer

I_req = 68.18 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 2.811 kN

Euler buckling: a slender column becomes unstable at P_cr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 50·25³/12 = 65104 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·65104/4000² = 2811 N = 2.811 kN

Answer

P_cr = 2.811 kN

14.Column buckling

Answer: σ_cr = 11.05 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 0.5·5000 = 2500 mm

KL/r = 2500/10 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/250² = 11.05 MPa

Answer

σ_cr = 11.05 MPa

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